If mean of 20 observation is 10 and standard deviation is 2. If one observation was 8 but misread as 12, then correct SD will be equal to

Options

A. `(3 sqrt (11))/5`
B. `sqrt (11)/5`
C. `(2 sqrt (11))/5`
D. `(99)/(25)`

Answer

To solve the problem step by step, we need to find the corrected standard deviation after one observation was misread.

### Step 1: Calculate the original sum of observations
Given that the mean of 20 observations is 10, we can find the total sum of the observations (Σxi):
\[ \text{Mean} = \frac{\Sigma xi}{n} \implies \Sigma xi = \text{Mean} \times n = 10 \times 20 = 200 \]

### Step 2: Identify the incorrect observation and correct it
The observation was misread as 12 instead of the actual value of 8. To correct the sum:
- Subtract the incorrect observation (12) and add the correct observation (8):
\[ \text{Corrected sum} = 200 - 12 + 8 = 196 \]

### Step 3: Calculate the original sum of squares
The standard deviation (SD) is given as 2. We can find the sum of squares of the original observations (Σxi²) using the formula for variance:
\[ \text{Variance} = \frac{\Sigma xi^2}{n} - \left(\frac{\Sigma xi}{n}\right)^2 \]
Given that SD = 2, the variance is:
\[ \sigma^2 = 2^2 = 4 \]
Thus,
\[ 4 = \frac{\Sigma xi^2}{20} - 10^2 \implies 4 = \frac{\Sigma xi^2}{20} - 100 \]
Rearranging gives:
\[ \frac{\Sigma xi^2}{20} = 104 \implies \Sigma xi^2 = 104 \times 20 = 2080 \]

### Step 4: Adjust the sum of squares for the corrected observation
Now we need to adjust the sum of squares for the corrected observation:
- Subtract the square of the incorrect observation (12² = 144) and add the square of the correct observation (8² = 64):
\[ \text{Corrected } \Sigma xi^2 = 2080 - 144 + 64 = 2000 \]

### Step 5: Calculate the new mean
The new mean after correction is:
\[ \text{New Mean} = \frac{\text{Corrected Sum}}{n} = \frac{196}{20} = 9.8 \]

### Step 6: Calculate the new standard deviation
Using the new sum of squares and the new mean, we can find the new variance:
\[ \text{New Variance} = \frac{\Sigma xi^2}{n} - \left(\text{New Mean}\right)^2 \]
Substituting the values:
\[ \text{New Variance} = \frac{2000}{20} - (9.8)^2 = 100 - 96.04 = 3.96 \]
Thus, the new standard deviation (SD) is:
\[ \text{New SD} = \sqrt{3.96} = \sqrt{\frac{99}{25}} = \frac{\sqrt{99}}{5} = \frac{3\sqrt{11}}{5} \]

### Final Answer
The corrected standard deviation is:
\[ \frac{3\sqrt{11}}{5} \]